Bathsheba has 5 paintings in her house, 1 each in a color palette of red, yellow, blue, green, and purple. She hangs them randomly in a horizontal order. What is the probability of hanging the blue painting first, and then the green?(1 point)

Responses

15+15
Start Fraction 1 over 5 End Fraction plus Start Fraction 1 over 5 End Fraction

15
Start Fraction 1 over 5 End Fraction

15⋅15
Start Fraction 1 over 5 End Fraction times Start Fraction 1 over 5 End Fraction

15⋅14

1 answer

To find the probability of hanging the blue painting first and then the green painting second from Bathsheba's collection of 5 paintings, we can follow these steps:

  1. Choice for the first painting: There are 5 paintings available. The probability of choosing the blue painting first is \( \frac{1}{5} \).

  2. Choice for the second painting: After hanging the blue painting first, there are now 4 paintings left (red, yellow, green, and purple). The probability of then choosing the green painting second is \( \frac{1}{4} \).

  3. Combined probability: The two choices are independent, so we multiply the probabilities: \[ P(\text{Blue first and Green second}) = P(\text{Blue first}) \times P(\text{Green second | Blue first}) = \frac{1}{5} \times \frac{1}{4} = \frac{1}{20}. \]

Thus, the answer is: \[ \frac{1}{20}. \]

This choice does not directly match any of the given responses. However, if we simplified it into fractions, we recognize that \(1/20\) could be interpreted through the alternatives listed.

If we match it to the responses you showed:

  • The closest option is "15⋅14," which might suggest something related to total arrangements but does not directly correlate as a probability. It's best to stick with calculated values.

In conclusion, the probability \( \frac{1}{20} \) for blue being first and green being second does not appear on the options given.